Test: TCP/ UDP & Sockets - Computer Science Engineering (CSE) MCQ

We need the maximum throughput, and for that we need to send as much data as possible to fully utilize the bandwidth.
so, maximum packets that can be sent = 1 + 2a = 9(after calculation) for 100% efficiency.
But we have a window size of 5 only, so we can send only 5 packets at max.
Efficiency = 5/9
Now, A/Q, Bandwidth of the channel (BW) = L/Tt = 1000 / (50*10^-6) = 20*10^6 bytes/sec.
So, max. throughput achievable = Efficiency * BW = 5/9 * 20*10^6 = 11.11 * 10^6 bytes/sec. (B)

C) M+N
Because Ws+Wr ≤ Sequence numbers (as the maximum number of unacknowledged packets at sender will be Ws and at the receiver it will be Wr, similar to the sequence numbering in Selective Repeat)
where Ws is size of sender window and Wr is receiver window's size.

Answer: B
Transmission TIme = 100*8 bits/20 Kbps = 40 ms
Propagation Time = 400 ms
Efficiency = Window Size*Transmission Time/(Transmission Time + 2*Propagation Time) = 10*40/(40+2*400) = .476
Maximum Data Rate = .476*20 Kbps = 9.52 Kbps which is close to option B.

In slow-start phase, for each ACK, the sender increases the current transmit window by Maximum Segment Size (MSS). In
the question it is given a packet consists of 2000 bytes and that can be taken as MSS. So, after two ACKs, current transmit
window
= 4000 + 2000 + 2000
= 8000

ans b)
In selective reject protocol, the maximum window size must be half the sequence number space = 2ⁿ/2 = 2^n-1.
For Go-back n, the maximum window size can be 2^n-1

Answer: B
Round Trip Time = 80ms
Frame size = 32
8 bits
Bandwidth = 128kbps
Transmission Time =

Let
n be the window size.
Utilization =

where

For maximum utilization:
n = 41 which is close to option (B).

Since all packets are ready initially itself, we can assume a time out is detected after all possible packets are sent. So, the sending happens as shown in figure (I draw the figure assuming 10 packets. For 9 packets answer will be 16).

In computer networks, goodput is the application level throughput, i.e. the number of useful information bits delivered by
the network to a certain destination per unit of time. (From wikipedia).
So, successful delivery of packet can be assured if ACK has been received for it.
So till time 'i' we would have transmitted 'i' packets but only (i-N) can be acknowledged as minimum time for a packet to
get Acknowledged is N (since RTT is N which is equal to the window size, there is no waiting time for the sender).
So successfully delivered packets = (i-N)
Time for transmission = i
Goodput = Successfully delivered data/ Time = (i-N)/i = 1- N/i
Therefore (A)

Answer: C

for maximum utilization η = 1
so,

gives the number of packets that are sent. It means that it is the Sender's Window Size.
But we also know that

so, to get the minimum number of bits needed to represent the sequence numbers we should consider what protocols are in use.
if GBN ARQ:

if Selective Repeat ARQ :

Distance from Station A to Satellite = 36504 x 10³ m

Time to reach satellite = 

Efficiency is the ratio of the amount of data sent to the maximum amount of data that could be sent. Let X be the packet size.
In Go-Back-N, within RTT we can sent n packets. So, useful data is nxX, where X is the packet size. Now, before we can sent another packet ACK must reach back. Time for this is transmission time for a packet (other packets are pipelined and we care only for first ACK), and RTT for a bit (propagation times for the packet + propagation time for ACK +
transmission time for ACK - neglected as per question)

Packet Size = 960 bits = 120 Bytes
so option (A)

Bandwidth won't be halved in full duplex. 
Propagation time is given as 25 ms.
Bandwidth = 10⁶ bps.
So, to fully utilize the channel, we must send 10⁶ bits into the channel in a second, which will be 1000 frames per second as each frame is 1000 bits. Now, since the propagation time is 25 ms, to fully pack the link we need to send at least 1000 * 25 * 10^-3 = 25 frames. So, we need ⌈log₂ 25⌉ = 5 bits.

A and C are not the answers as the second byte of IP differs and subnet mast has 255 for second byte.
Consider B, (& for bitwise AND)
10.35.28.2 & 255.255.31.0 = 10.35.28.0 (28 = 111002)
10.35.29.4 & 255.255.31.0 = 10.35.29.0 (29 = 111112)
So, we get different subnet numbers
Consider D.
128.8.129.43 & 255.255.31.0 = 128.8.1.0 (129 = 100000012)
128.8.161.55 & 255.255.31.0 = 128.8.1.0 (161 = 101000012)
The subnet number matches. So, D is the answer.

The Answer must be A.
For 1st packet,
(128.75.43.16) && (255.255.255.0) = (128.75.43.0) since {16 && 0 = 0}, as well as
(128.75.43.16) && (255.255.255.128) = (128.75.43.0) since {16 && 128 = 0}.
Now, since both these subnet masks are producing the same Network ID, hence The one with greater number of ones will
be selected, and the packet will be forwarded there. Hence packet 1 will be forwarded to Eth1.
For 2nd packet,
(192.12.17.10) when anded with each of the subnet masks does not match with any of the network ID, since:
(192.12.17.10) && (255.255.255.0) = (192.12.17.0) {Does not match with any of the network addresses}
(192.12.17.10) && (255.255.255.128) = (192.12.17.0) {Does not match with any of the network addresses}
(192.12.17.10) && (255.255.255.255) = (192.12.17.10) {Does not match with any of the network addresses}
Hence, Default interface must be selected for packet 2, i.e Interface Eth2.

C is answer since you have 6 zeroes so u can make 64-2 hosts

D is correct answer.

to form subnet for 64 departments we need 6 continuous bit and the value of 11111100 = 252. organization has class B network so subnet mask would be 255.255.252.0

Here
C) Seems correct answer.
Say you type www.google.com
FIrst you send DNS request to get IP address. Then you establish connection with IP of google using TCP. Finally you start
talking in HTTP !

Option B: "Sequence numbers allow receivers to discard duplicate packets and properly sequence reordered packets."
Option C: "When congestion is detected, the transmitter decreases the transmission rate by a multiplicative factor; for example, cut the congestion window in half after loss." (Additive Increase/multiplicative
decrease)
Option D: "Acknowledgments allow senders to determine when to retransmit lost packets."
So, A is answer.

a) it should be byte

Initially token bucket is full.
Rate at which it is emptying is (20-10) MBps.
Time taken to empty token bucket of 1 MB is 1/10 i.e 0.1 sec.
Data send in this time is 0.1 * 20 = 2 MB (rate at which bucket is emptying is different from rate at which data is send) .
Data left to send is 12 - 2 = 10 MB .
Now bucket is empty and rate of token arriving is less than that of going out so effective data speed will be 10 MBps.
Time to send remaining 10 MB will be 1 sec. So total time is 1+0.1= 1.1sec

New tokens are added at the rate of r bits/sec which is
2Mbps in the given question.
Capacity of the token bucket (b) = 16 Mbits
Maximum possible transmission rate (M) = 10Mbps
So the maximum burst time = b/(M-r) = 16/(10-2) = 2 seconds
here is the animation for token bucket hope this will help us to understand the concept.